1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity = 10 m/s2. Find (a) The acceleration of the system (b) The tension in the cord!
Solution
Known :
Mass of the box 1 (m1) = 2 kg
Mass of the box 2 (m2) = 3 kg
Acceleration due to gravity (g) = 10 m/s2
Weight of the box 1 (w1) = m1 g = (2)(10) = 20 Newton
Weight of the box 2 (w2) = m2 g = (3)(10) = 30 Newton
Solution :
(a) magnitude and direction of the acceleration
w2 > w1 so the box 2 accelerates downward and the box 1 accelerates upward.
Forces that has the same direction with acceleration (w2 and T1), its sign positive. Forces that has opposite direction with acceleration (T2 and w1), its sign negative.
∑F = m a
w2 – T2 + T1 – w1 = (m1 + m2) a ——-> T1 = T2 = T
w2 – T + T – w1 = (m1 + m2) a
w2 – w1 = (m1 + m2) a
30 – 20 = (2 + 3) a
10 = 5 a
a = 10 / 5
a = 2 m/s2
Magnitude of the acceleration is 2 m/s2.
(b) The tension force
The box 2 :
There are two forces acts on the box 2 : first, weight of the box 2 (w2), points downward so it’s positive. Second, tension force exerted on the box 2 (T2), points upward so it’s negative. Apply Newton’s second law of motion.
∑F = m a
w2 – T2 = m2 a
30 – T2 = (3)(2)
30 – T2 = 6
T2 = 30 – 6
T2 = 24 Newton
Box 1 :
There are two forces acts on the box 1. First, weight of the box 1 (w1), points downward so it’s negative. Second, the tension force exerted on the box 1 (T1) points upward so it’s positive. Apply Newton’s second law of motion :
∑F = m a
T1 – w1 = m1 a
T1 – 20 = (2)(2)
T1 – 20 = 4
T1 = 20 + 4
T1 = 24 Newton
Magnitude of the tension force = T1 = T2 = T = 24 Newton
2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m/s2, coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!
Solution
Known :
Mass of the object 1 (m1) = 2 kg
Mass of the object 2 (m2) = 4 kg
Acceleration due to gravity (g) = 10 m/s2
Coefficient of the static friction (μs) = 0.4
The coefficient of the kinetic friction (μk) = 0.3
Weight of the object 1 (w1) = m1 g = (2)(10) = 20 Newton
Weight of the object 2 (w2) = m2 g = (4)(10) = 40 Newton
Normal force exerted on the object 1 (N) = w1 = 20 Newton
Force of the static friction exerted on the object 1 (fs) = μs N = (0.4)(20) = 8 Newton
Force of the kinetic friction exerted on the object 1 (fk) = μk N = (0.3)(20) = 6 Newton
Wanted: acceleration (a)
Solution :
w2 > fs (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (fk). Apply Newton’s second law of motion :
∑F = m a
w2 – the = (m1 + m2) a
40 – 6 = (2 + 4) a
34 = 6 a
a = 34 / 6 = 17 / 3
a = 5.7 m/s2
Magnitude of the acceleration = 5.7 m/s2
[wpdm_package id=’484′]
- Mass and weight
- Normal force
- Newton’s second law of motion
- Friction force
- Motion on horizontal surface without friction force
- The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
- Motion on the inclined plane without friction force
- Motion on the rough inclined plane with the friction force
- Motion in an elevator
- The motion of bodies connected by cord and pulley
- Two bodies with the same magnitude of accelerations
- Rounding a flat curve – dynamics of circular motion
- Rounding a banked curve – dynamics of circular motion
- Uniform motion in a horizontal circle
- Centripetal force in uniform circular motion
